We all know plenty of examples of multilinear forms in finitely many variables (e.g. determinants). However, I am missing an interesting example of a form in infinitely many variables, linear in each. The important point here is the word "interesting" ; the example should occur naturally in some nice situation in algebra, or geometry, or analysis... (One can easily find non-interesting examples.) I thought that maybe regularized determinants could provide examples, but I don't know much about these. Thanks for ideas !

3 Answers
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Consider the Hilbert space $H=L^2(S^1, \mathbf{C}^n)$ and the subspaces $H_+, H_-$ of positive and negative Fourier modes. One can construct the Hilbert space of infinite wedge products $V=\bigwedge(H_+)\hat{\otimes}\bigwedge(H_-)^*$.

Just as $H$ carries an action of $LSU(n)=Maps(S^1, SU(n))$ by pointwise multiplication, $V$ carries an action of the central extension $\widetilde{LSU(n)}$. It turns out that all irreducible positive-energy level 1 representations occur as summands in $V$.

I'm having some difficulty parsing your second paragraph. Do you know of any references where these notions are made more explicit?
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AlexArvanitakisAug 30 '12 at 17:12

Whoops, I'm sorry but I know very few Fourier analysis. What are the positive and negative modes ? What are "irreducible positive-energy level 1 representations " ? How does that answer the question ?
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Matthieu RomagnyAug 30 '12 at 19:49

@Matthieu, if $S^1\subset\mathbf{CP}^1$ is the unit circle, $H_+$ is the subspace of maps that extend to the unit disk (Hardy space), $H_-$ is the subspace of maps that extend to its complement. "Positive-energy" condition is just a certain finiteness assumption; level refers to the action of the central element. This is all discussed in chapter 10 of Pressley-Segal's "Loop Groups". How this relates to the question: $V$ is the space of multilinear forms in infinitely-many variables, which can be thought of as an analog of the space of volume forms in the absence of the notion of "top" forms.
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Pavel SafronovAug 30 '12 at 23:33

Hmmm... I think that the functional determinants (as in http://en.wikipedia.org/wiki/Functional_determinant) of Quantum Mechanics and Quantum Field Theory appear rather naturally. These CAN be defined rigorously, insofar as the defining Feynman path integrals are, in this case, defined rigorously. The most readable introduction to rigorous path integration I've read is "A Modern Approach to Functional Integration" by John R. Klauder.

EDIT: I did some reading. The bosonic path integral expression (as in the wikipedia page I linked to earlier) for the functional determinant may fail to be multilinear (though it is rigorous), whereas the fermionic path integral expression

should be genuinely multilinear and valid for every "reasonable" S (including nonsymmetric/nonhermitian ones) IF fermionic functional integrals in infinite variables can be consistently defined to yield sensible results (e.g $\int 1 \mathcal D c=0$), which, as far as I know, has not yet been done rigorously.

One point I'm wondering about is whether these determinants are indeed multilinear in any sense.
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Matthieu RomagnyAug 30 '12 at 19:41

You know, I never thought about explicitly proving that. I think I can show it formally, in the sense that if one writes the operator $S$ (as in the wikipedia page) as a sum of terms $|v_i\rangle\langle i|$ where the $\langle i|$ form a basis, then the functional determinant is linear wrt every $|v_i\rangle$ --that is, the columns. I'm going to sleep on it.
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AlexArvanitakisAug 31 '12 at 2:35

@Delio Mugnolo: If you fix a basis, you can regard a linear function on an infinite dimensional space as an infinite-variable mapping of the corresponding one-dimensional spaces, but the mapping will not be multilinear: you will not have $f(x_1+y_1,x_2,x_3,\dots)=f(x_1,x_2,x_3,\dots)+f(y_1,x_2,x_3,\dots)$.
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Emil JeřábekAug 30 '12 at 14:02

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I see. I did not understand this linear algebraic definition.
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Delio MugnoloAug 30 '12 at 14:23

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So, perhaps the TITLE should say multilinear and not linear ... ???
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Gerald EdgarSep 2 '12 at 22:34